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2 years ago

I need help with number 5 :( pls help

Mathematics

1 answer:

jok3333 [9.3K]2 years ago

7 0

Answer:

the answer is c

Step-by-step explanation:

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Can someone help me in this pretty please and asap!!! ;((

zepelin [54]

Answer:

(x -5)² + (y +4)² = 100 Should be the correct answer, hope this helps :)

Step-by-step explanation:

A circle centered at (h, k) with radius r will have equation ...

... (x -h)² + (y-k)² = r²

The point satisfies the equation for the circle. Filling in the given numbers, we have ...

... (x -5)² + (y+4)² = (-3-5)² + (2+4)² . . . . . . (h, k) = (5, -4), (x, y) = (-3, 2)

... (x -5)² + (y +4)² = 64 +36

6 0

3 years ago

2y-x=2 substitute solution?

DerKrebs [107]

Answer:

Y=1 X=0

Step-by-step explanation:

5 0

3 years ago

An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then

Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

$\angle Q=110^\circ$

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

$q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km$

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

$=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)$

(b)Flight Direction of Y

Using Law of Sines

$\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)$